In a stunt scene for a new action movie a motorcyclist will take off horizontall

Question

In a stunt scene for a new action movie a motorcyclist will take off horizontally from the top of one building, travel across a gap and then land on the roof of a lower building next to it. The two buildings are separated horizontally by a gap of 12 m and the roof level of the lower building is 8 m below that of the first building. a) If the motorcycle has just enough speed to make the jump safely, how long does it take for the motorcycle to make the jump between the two buildings? b) What is the minimum horizontal take-off speed that the motorcycle must have to make the jump safely?

Explanation / Answer

find time it takes to drop 8 m: d = (vo)(t) + 0.5at^2 where d is the vertical displacement (d = -8) where vo is the initial vertical velocity (vo = 0) where t is the time (are solving for this) where a is the vertical acceleration (a = -9.8 due to gravity) -8 = 0.5(-9.8)(t^2) t = 1.278 seconds find the horizontal speed to travel the horizontal distance of 12 m during the time interval found above: d = (vo)(t) + 0.5at^2 where d is the horizontal displacement (d = 12) where vo is the initial horizontal speed (solving for this) where t is the time (t = 1.278 from part a) where a is the horizontal acceleration (a = 0) 12 = (vo)(1.2778) vo = 9.39 m/s bol

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